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Proof step 2 from proof ri-inj-rem (previous step)

ri-inj(ri) ⊦ ri-inj(ri \ ks)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ri[key] = ri[key0], key ∈ ri, key0 ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key = key0, key ∈ ks, key0 ∈ ks

Rule argument

Additional information

Used simplifier rules: ri-inj: ⊦ ri-inj(ri) ↔ (∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1);% ⊦ ¬ key ∈ ks → (ri \ ks)[key] = ri[key]; (s)% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)